Timeline for Approximating a convex disk by an ellipse
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2017 at 0:24 | comment | added | Iosif Pinelis | I still don't understand what the substantial innovation here is. Is it to use Euler--Lagrange eqs.? How can you get them for such a non-smooth function as $|\cdot|$? And if you can get them and use them to prove that $K$ is unique, why do you still need to prove an inequality for that $K$? In addition, the claim "areas like $|F \cap D|$ can be written as $\frac{1}{2}\int \min(f,r_D)^2$" is incorrect -- in general, the expression is rather more complicated. | |
| Jan 4, 2017 at 20:52 | comment | added | user44143 | @IosifPinelis, in your answer the inequality of $\delta$'s is supposed to be proved for all K; in that sense, it involves an infinite-dimensional search space. The intended advantage of my approach is that the Euler-Lagrange equations determine a unique K, and then it suffices to prove an inequality for that K. | |
| Jan 4, 2017 at 20:47 | comment | added | user44143 | @WlodekKuperberg, I never actually have to calculate those transformations, I just work in the coordinates they define | |
| Jan 4, 2017 at 20:39 | comment | added | Iosif Pinelis | @MattF. : In my answer, one also had to deal with just three parameters ($t$ and the two coordinates of the vector $b$), in addition to a function describing the arbitrary convex set. However, there one had to deal just with an arithmetic mean, rather than with both the arithmetic and geometric ones. | |
| Jan 4, 2017 at 20:32 | comment | added | Iosif Pinelis | @WlodekKuperberg : Such an affine transformation was rather explicitly constructed in my answer. | |
| Jan 4, 2017 at 19:20 | comment | added | Wlodek Kuperberg | @MattF., if it helps with computations: For every pair of ellipsoids of equal volume there exists an affine transformation under which the images of the two ellipsoids are congruent. | |
| Jan 4, 2017 at 15:45 | comment | added | user44143 | @IosifPinelis, we have an ellipse with axes (1,1) and an ellipse with axes (a, 1/a). We need an ellipse with axes (b, 1/b) to keep the area constant. What other choice of b would be reasonable? This one treats the two ellipses symmetrically: if C,D,E are of the above form with a,h,k; and C',D',E' are of the above form with a',h',k'; and E'=MC, C'=ME, then also D'=MD. I think this is the only choice with that symmetry. | |
| Jan 4, 2017 at 14:49 | comment | added | Iosif Pinelis | Why the geometric means? Is there any reason to believe that they are what is needed? | |
| Jan 4, 2017 at 5:24 | history | answered | user44143 | CC BY-SA 3.0 |